如何使用Python计算目录中的文件数量

I need to count the number of files in a directory using Python. I guess the easiest way is len(glob.glob('*')), but that also counts the directory as a file.

我需要使用Python计算一个目录中文件的数量。我想最简单的方法是len(glob.glob('*'))，但它也将目录作为一个文件。

Is there any way to count only the files in a directory?

有没有办法只计算目录中的文件?

18 个解决方案

#1

os.listdir() will be slightly more efficient than using glob.glob. To test if a filename is an ordinary file (and not a directory or other entity), use os.path.isfile():

os.listdir()将比使用glob.glob稍微高效一些。要测试文件名是否为普通文件(而不是目录或其他实体)，请使用os.path.isfile():

import os, os.path

# simple version for working with CWD
print len([name for name in os.listdir('.') if os.path.isfile(name)])

# path joining version for other paths
DIR = '/tmp'
print len([name for name in os.listdir(DIR) if os.path.isfile(os.path.join(DIR, name))])

#2

import os

path, dirs, files = next(os.walk("/usr/lib"))
file_count = len(files)

#3

For all kind of files, subdirectories included:

对于所有类型的文件，包括子目录:

import os

list = os.listdir(dir) # dir is your directory path
number_files = len(list)
print number_files

Only files (avoiding subdirectories):

只有文件(避免子目录):

import os

onlyfiles = next(os.walk(dir))[2] #dir is your directory path as string
print len(onlyfiles)

#4

This is where fnmatch comes very handy:

这就是fnmatch的用武之地:

import fnmatch

print len(fnmatch.filter(os.listdir(dirpath), '*.txt'))

More details: http://docs.python.org/2/library/fnmatch.html

更多信息:http://docs.python.org/2/library/fnmatch.html。

#5

def directory(path,extension):
  list_dir = []
  list_dir = os.listdir(path)
  count = 0
  for file in list_dir:
    if file.endswith(extension): # eg: '.txt'
      count += 1
  return count

#6

import os
print len(os.listdir(os.getcwd()))

#7

This uses os.listdir and works for any directory:

它使用操作系统。listdir，适用于任何目录:

import os
directory = 'mydirpath'

number_of_files = len([item for item in os.listdir(directory) if os.path.isfile(os.path.join(directory, item))])

this can be simplified with a generator and made a little bit faster with:

这可以用一个发电机简化，并且用:

import os
isfile = os.path.isfile
join = os.path.join

directory = 'mydirpath'
number_of_files = sum(1 for item in os.listdir(directory) if isfile(join(directory, item)))

#8

If you want to count all files in the directory - including files in subdirectories, the most pythonic way is:

如果要计算目录中的所有文件——包括子目录中的文件，最符合python的方法是:

import os

file_count = sum(len(files) for _, _, files in os.walk(r'C:\Dropbox'))
print(file_count)

We use sum that is faster than explicitly adding the file counts (timings pending)

我们使用sum，它比显式地添加文件计数(计时待定)要快

#9

def count_em(valid_path):
   x = 0
   for root, dirs, files in os.walk(valid_path):
       for f in files:
            x = x+1
print "There are", x, "files in this directory."
return x

Taked from this post

采取从这篇文章

#10

import os

def count_files(in_directory):
    joiner= (in_directory + os.path.sep).__add__
    return sum(
        os.path.isfile(filename)
        for filename
        in map(joiner, os.listdir(in_directory))
    )

>>> count_files("/usr/lib")
1797
>>> len(os.listdir("/usr/lib"))
2049

#11

I am surprised that nobody mentioned os.scandir:

我很惊讶没有人提到os.scandir:

def count_files(dir):
    return len([1 for x in list(os.scandir(dir)) if x.is_file()])

#12

Luke's code reformat.

卢克的代码重新格式化。

import os

print len(os.walk('/usr/lib').next()[2])

#13

Here is a simple one-line command that I found useful:

这里有一个简单的单行命令，我觉得很有用:

print int(os.popen("ls | wc -l").read())

#14

import os

total_con=os.listdir('<directory path>')

files=[]

for f_n in total_con:
   if os.path.isfile(f_n):
     files.append(f_n)


print len(files)

#15

If you'll be using the standard shell of the operating system, you can get the result much faster rather than using pure pythonic way.

如果您将使用操作系统的标准shell，您可以更快地得到结果，而不是使用纯python方式。

Example for Windows:

窗口的示例:

import os
import subprocess

def get_num_files(path):
    cmd = 'DIR \"%s\" /A-D /B /S | FIND /C /V ""' % path
    return int(subprocess.check_output(cmd, shell=True))

#16

I found another answer which may be correct as accepted answer.

我找到了另一个可能是正确的答案。

for root, dirs, files in os.walk(input_path):    
for name in files:
    if os.path.splitext(name)[1] == '.TXT' or os.path.splitext(name)[1] == '.txt':
        datafiles.append(os.path.join(root,name)) 


print len(files) 

#17

I used glob.iglob for a directory structure similar to

我用一团。用于类似于的目录结构的iglob

data
└───train
│   └───subfolder1
│   |   │   file111.png
│   |   │   file112.png
│   |   │   ...
│   |
│   └───subfolder2
│       │   file121.png
│       │   file122.png
│       │   ...
└───test
    │   file221.png
    │   file222.png

Both of the following options return 4 (as expected, i.e. does not count the subfolders themselves)

以下两个选项都返回4(如预期的那样，即不计算子文件夹本身)

• len(list(glob.iglob("data/train/*/*.png", recursive=True)))
• len(列表(glob.iglob(“数据/火车/ * / *。png”,递归= True)))
• sum(1 for i in glob.iglob("data/train/*/*.png"))
• sum(1 for i in glob.iglob("data/train/*/*.png"))

#18

i did this and this returned the number of files in the folder(Attack_Data)...this works fine.

我这样做了，它返回了文件夹(Attack_Data)中的文件数量……这是很好。

import os
def fcount(path):
    #Counts the number of files in a directory
    count = 0
    for f in os.listdir(path):
        if os.path.isfile(os.path.join(path, f)):
            count += 1

    return count
path = r"C:\Users\EE EKORO\Desktop\Attack_Data" #Read files in folder
print (fcount(path))